Question

An old-fashioned vinyl record rotates on a turntable at 45 rpm. What is the angular speed in rad/s?

What is the period of motion

Answer 1

3

Answer 2

Concepts and reason

The main concepts required to solve this problem are the angular velocity and period of motion.

Initially, write the equation for the relation of the angular velocity and the period of motion. Use the conversion method and calculate the angular speed of the old-fashioned vinyl and find the period of motion by using the equation for the relation of angular speed and the period of motion.

Fundamentals

Angular speed can be defined as the angular distance rotated by an object in a circular path per unit time.

The equation for the relation of the angular speed and the period of motion is,

$\omega = \frac{{2\pi }}{T}$

Here, $\omega$ is the angular speed, and T is the period of motion.

The full form of ${\rm{rmp}}$ is revolutions per minutes which can be written as rev/min.

The conversion of rev into rad and min into s are as follows,

$1{\rm{ rev}} = 2\pi {\rm{ rad}}$

And

${\rm{1 min}} = 60{\rm{ s}}$

The angular speed of old-fashioned vinyl is,

$\begin{array}{c}\\\omega = 45{\rm{ rev/min}}\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)\left( {\frac{{{\rm{1 min}}}}{{60{\rm{ s}}}}} \right)\\\\ = 45{\rm{ rev/min}}\left( {\frac{{2\left( {3.14} \right){\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)\left( {\frac{{{\rm{1 min}}}}{{60{\rm{ s}}}}} \right)\\\\ = 4.71{\rm{ rad/s}}\\\end{array}$

The equation for the period of motion is,

$\omega = \frac{{2\pi }}{T}$

Here, $\omega$ is the angular speed of the old-fashioned vinyl and T is the time-period of the motion

Rearrange the above equation for T.

$T = \frac{{2\pi }}{\omega }$

Substitute 4.71 rad/s for $\omega$ , and 3.14 for $\pi$ in above equation.

$\begin{array}{c}\\T = \frac{{2\left( {3.14} \right)}}{{\left( {4.71{\rm{ rad/s}}} \right)}}\\\\ = 1.33{\rm{ s}}\\\end{array}$

Ans:

The angular speed of the old-fashioned vinyl is 4.71 rad/s.